5

D dt dx 12 +3...

Question

D dt dx 12 +3

d dt dx 12 +3



Answers

$\int_{0}^{\infty} x^{3} e^{-x} d x$ is (a) 2 (b) 6 (c) 24 (d) 12

Hello. So we're trying to evaluate into girl of t E to a native three tt This is a standard integration by parts problem. So you have to figure out a you and a devi to decompose the problem with. Usually you want to find a you that you khun keep differentiating until you get to like one. So that you, Khun, basically get rid of it. And the DV needs to be something that you can integrate visioning your conventional methods. So in this case, it's easy to see the U equals T and D vehicles even Eva three t I say easy because I've done this so many times. It comes with experience. Do you would then be equal to DT? The is equal to e to the negative one third e to the negative three t So what happens next? Put it together. I color coded it. So it's easier to see I put the constants in France. It's easier to work with afterwards, So you have equals to you tempt me which is they won thirty to the in the eternity of three t minus then ago VD you So that's minus ah native one third in a girl of eighteen eighty three T d. T. So you get a plus one third in the next line. Um, and then integrating into ninety three t by itself would just would he also give you another native one third term in a man we get eked native one thirty, eatin a three T minus one ninth eater native three t plus C.

We have to determine whether the following claim is true false. The integral from one to e cubed of one over T. D. T. Is equal to three. So let's actually calculate the center girl. So the first thing to note is that the integral of one over X. Dx is the natural log of the absolute value of X. Okay, so by definition this is simply going to be the natural log of the absolute value of T. Evaluated from T equals 12 T equals E cubed. Ok, so what you do is you plug in the upper bound, this guy right here and this is going to be the natural log of he cubed minus the natural log of this lower bound right here at T equals one. So plug in one right there. Now the natural log of one is zero. Because remember the natural log function is asking what number do I have to raise? E to either the what to get one by definition, regardless of what the base is. The only thing that will uh be equal to zero is if I plug in zero, I'm sorry. One each of the zeros one. So the natural log of one has to be zero. So this essentially since we're subtracting it doesn't matter. It goes away and we're left with um the natural log of acute now is a positive number. And since you're multiplying a positive number by itself, a total of three times the absolute value symbols are actually unnecessary. And for this you're asking what power do I need to raise E to eat of the what we'll get me? Cute. That's a pretty easy question. And the answer is just three. Making this evaluated to three, verifying the claim as true.

We know that you is to to the t plus three, which means that de you is to the g natural over to D. T. Which means that u to the t dtv as do you over natural too. Which means we pull out one over natural with two times the integral one over you, do you? Which gives us one over Natural of two. Tim's Natural Look of You plus C, which gives a snatcher walk of tea to the two to the teeth plus three. Divide by natural of two, plus C is our solution.

Right. Um They're asking you to find the integral uh I guess the in my opinion, they want you to do the integral of F. A bex dx first and then find the derivative of that. So let's talk about how um there's a function out there who's derivative is equal to Alfa Becks. Um So what we're trying to get at then is that the integral of F of X could be represented as, you know, F prime of X. Capital of I'm sorry, I said that wrong just F of X because it's the anti derivative, but it could have any constant. So plus C. But as soon as you're asked to find the derivative of that, well, the derivative of F is equal to how I've defined it. F. A bex and the derivative of a constant would be zero, which is simply equal to F A bex. Um So this is the answer that they're looking for, based off of how the derivative and the integral undo each other, so to speak. Um So there you go.


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